Question #aa375

Question

2057 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-05T12:55:15

$x = \frac{1}{8} .$ $x=1/8.$

Recall the Defn. of $a r c cos$ $arc cos$ function :

$a r c cos x = θ, | x | \leq 1 \Leftrightarrow cos θ = x, θ \in [0, π] .$ $arc cos x=theta , |x| <=1 iff costheta=x, theta in [0,pi].$

Let, $theta=arc cos (3/4)., :. costheta=3/4, & 0<=theta<=pi.$

$because costheta=3/4 > 0, :. 0 <=theta<=pi/2.$

Now, given that, $arc cos x=2arc cos (3/4)=2theta.$

$:. cos2theta=x;" but, "cos2theta=2cos^2theta-1=2(3/4)^2-1.$

$:. x=2(3/4)^2-1=1/8.$

Answer 2 · 2017-07-06T04:38:41

cos x = 0.125
arc $x = +- 82^@82$

arccos x = 2 arccos (3/4)
Calculator gives
$cos t = 3/4$ --> arc $t = +- 41^@41$
$2t = +- 82^@82$
arc $x = +- 82^@82$ --> $cos x = 0.125 = 1/8$